I'm having trouble solving the following problem: Find the coefficients $a_1,a_2,a_3$ so that the signal $x[n] = Ar^n \sin(\Omega_0 n)u[n]$, $|r|<1$, satisfies the following difference equation for $n \geq 3$:

\begin{equation} x[n]-a_1 x[n-1]-a_2 x[n-2]-a_3 x[n-3]=0 . \quad n \geq 3 . \end{equation}

My effort:

Since the Z-transform at a first look seems complicated, I thought of moving to the z-domain (I'm not sure what Z-transform should be used here, the one-sided one or the bilateral), I get: \begin{align*} X(z) - a_1z^{-1}X(z) - a_2z^{-2}X(z) - a_3z^{-3}X(z)= 0 \\ X(z)(1 - a_1z^{-1} - a_2z^{-2}-a_3z^{-3})=0 \tag{$X(z) \neq 0$} \\ 1-a_1z^{-1}-a_2z^{-2}-a_3z^{-3}=0 \\ z^3 - a_1z^{2} - a_2z - a_3 = 0 \end{align*} I'm stuck here since third degree polynomial equations do not have an analytical solution. I will share that I'm given the following 4 possible answers: \begin{aligned} & a_1=r \sin \left(\Omega_o\right), \; a_2=r^2,\; a_3=1 . \\ & a_1=2 r \sin \left(\Omega_o\right), \; a_2=-r^2,\; a_3=1 . \\ & a_1=2 r \cos \left(\Omega_o\right),\; a_2=-r^2 ,\; a_3=0 . \\ & a_1=A,\; a_2=r, \; a_3=\Omega_o . \end{aligned}

Any help would be greatly appreciated! I feel like I'm missing something huge here. It seems like plugging in for $x[n]$ would make things unnecessary complex.


1 Answer 1


Here is one possible solution. Recognise that the finite difference equation is a convolution between the signal and a finite impulse response filter.

The z transform of x[n] is: $$x[z]=\frac{Arz^{-1}\sin(\Omega_0)}{1-2rz^{-1}\cos(\Omega_0)+r^2z^{-2}}$$

The z transform of the filter is: $$1-a_1z^{-1}-a_2z^{-2}-a_3z^{-3}$$

The time domain convolution becomes z domain multiplication: $$\frac{Arz^{-1}\sin(\Omega_0)}{1-2rz^{-1}\cos(\Omega_0)+r^2z^{-2}}\times (1-a_1z^{-1}-a_2z^{-2}-a_3z^{-3})$$

If the filter could cancel all the poles to give $Arz^{-1}\sin(\Omega_0)$, then the inverse transform will be an impulse delayed by one sample: $$Ar\sin(\Omega_0)\delta[n-1]$$ This is zero when $n\neq1$, which satisfies that your original difference equation is zero when $n\geq3$.

Therefore, simply equate coefficients of the following: $$1-a_1z^{-1}-a_2z^{-2}-a_3z^{-3}$$ $$1-2rz^{-1}\cos(\Omega_0)+r^2z^{-2}$$ This gives $a_1=2r\cos(\Omega_0)$, $a_2=-r^2$, and $a_3=0$, thus the third answer is correct.


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